The genes for ruby eyes (rb), tan body (t) and cut wings (ct) are all found on t

Question

The genes for ruby eyes (rb), tan body (t) and cut wings (ct) are all found on the X-chromosome of Drosophila melanogaster. All of these are recessive traits. They map in the order rb, ct, t with 15.5 map units between rb and ct and 5.5 map units between ct and t. Suppose you cross a cut wing male with a homozygous female that is both tan and has ruby eyes.
1) What will the F1 females look like? (Explain in detail)

2) Draw map of the section of the X chromosomes that has these 3 genes for the F1 females

3) Assuming you testcross your F1 females. What progeny classes would you expect?

Please explain answer in detail, I do not seek answer only but explaination to learn from.

Explanation / Answer

1. F1 females

Homozygous recessive 3 genes female - t---------ct-------rb    is crossed with

cut wing recessive gene and ruby eyes (rb), tan body (t) heterozygous male - t - ---------ct---------rb-      

F1 phenotype is t ----------------ct-----------------------rb-    

The normal eye and body from father is expressed and the recessive cut wing is expressed, since all the traits are recessive, both the recessive allele from father and mother is required to get expressed.

2) Draw map of the section of the X chromosomes that has these 3 genes for the F1 females

F1 phenotype

X chromosome   = t ----------------ct-----------------------rb-    

                                 5.5mu                   15.5mu

3.

t ----------------ct-----------------------rb- x t ----------------ct-----------------------rb-  

3) Assuming you testcross your F1 females. What progeny classes would you expect?

Expected Progeny

t ----------------ct-----------------------rb-  

100%    Normal eye and body with cut wing offspring

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